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\begin{document}
\title{Solving a Quartic Equation}
\author{Amit Yaron}
\maketitle
\begin{abstract}
Another one from YouTube, this time in Russian. But I don't understand
the language, and I feel that the video is too slow.
\end{abstract}

\paragraph{So, now let us solve $(x+3)^{4}+(x+5)^{4}=2$}

Everyone knows that $1+1=2$, one solution is really:

\begin{align*}
\begin{cases}
(x+3)^{4} & =1\\
(x+5)^{4} & =1
\end{cases} & \Rightarrow\\
\Rightarrow x & =-4
\end{align*}

But let us use a substitution to solve it more elegantly; I like using
the average of $3$ and $5$ to solve the equation,

so let us substitute:

$t=x+4$

and see terms cancel out, so a quartic equation will lool like a quadratic
one:

\[
(t-1)^{4}+(t+1)^{4}=2
\]

Let us expand the 2 terms:

\begin{align*}
(t^{4}-4t^{3}+6t^{2}-4t+1)+(t^{4}+4t^{3}+6t^{2}+4t+1) & =2\Rightarrow\\
\Rightarrow & 2t^{4}+12t^{2}+2=2\Rightarrow\\
\Rightarrow & 2t^{4}+12t^{2}=0\Rightarrow\\
\Rightarrow & 2t^{2}(t^{2}+6)=0
\end{align*}

So, one solution is:

\[
t=0\Rightarrow x=t-4=-4
\]

The other solutions are not real:

\[
t^{2}=\pm6\Rightarrow t=\pm\sqrt{6}i\Rightarrow x=t-4=-4\pm\sqrt{6}i
\]

\end{document}